Concepts in calculating diffusion

[MightyQuinn]

Hello experts in diffusion,

I am having some trouble understanding some concepts in calculating diffusion. First, I understand activity (or chemical potential) drives diffusion, but where does this come into play in the calculations? Is it in the activation energy (Q) in the Arrhenius equation (D=Doexp(-Q/RT))?
Next, where can I find data online for Do and Q for materials like 4340 steel etc? Or would I just use values for Carbon in FCC (Fe) if my temperature is austenitizing? I have found some values online which do not compare at all, so I assume they are wrong!

My reason is that I want to show in calculations that a steel say 4340 (with ~no Si content) will not carburize as much as a similar steel with ~3% Si or greater. How can I show this in calculations? Its obvious that it is because of the chemical potential gradient driving C away from the Si, but how can this be shown mathimatically. I have a good bit of literature and have done several searches and I cannot not find anything explaining this very important problem.

Any help is greatly appreciated

 

[uby]

I am not a metallurgist, but I am certain that the effect of solutes such as Si on the carburization of steel have been well studied in the literature. Rather than re-invent the wheel, I would look into the work of others first: I believe Mats Hillert published quite a bit on the subject over the years, and would be a good place to start.

As for your question on diffusion: Fick's law is properly written utilizing the chemical potential gradient rather than concentration gradient. Uphill diffusion against a concentration gradient (spinodal decomposition, liquid-phase solute extraction, etc.) could never occur if it weren't for the chemical potential gradient. The concentration gradient in many cases is the dominant factor, and is often used in Fick's law with satisfactory results.

 

[MightyQuinn]

Sorry, I just now got your reply. Thanks for your responce. I found some interesting documentation on the subject. And I do agree with your responce and have found the solution. If anyone is interested in more, let me know.